Is the norm of projections onto convex sets a convave function?

Question

Let $X\subset R^n$ be a closed convex set, and denote the projections of $x$ onto $X$ as $[x]^+$, i.e., $[x]^+ = argmin_z \{\|x-z\|_2 : z\in X\}$. Let $x\in X,0\neq z\in R^n$ be be given points and define $f:[0,\infty) \rightarrow R$ as $f(t) = \|[x+tz]^+ - x\|_2$. I wonder whether $f$ is a concave function in $t$?
I have ploted some figures of $f$ when $X \subset R^2$ are disks, rectangles and triangles, and in these examples $f$ is indeed a concave function. So I want to know whether $f$ is concave in general casess.
Here are some simplifications I made.
Using some translation we can assume that $x=0\in X$, then $f=\|[tz]^+\|$. And by definition $f$ is concave if the following inequality
$|[(\lambda t_1 + (1-\lambda t_2)) z]^+\|\geq \lambda\|t_1z]^+\|+(1-\lambda)\|[t_2 z]^+\|$
holds for all $t_1,t_2\geq 0$ and $\lambda\in [0,1]$.
But I don't how to preceed then.
And I know $f(t)/t$ is a monotonically nonincreasing, which may be useful for the proof.
Another closed relation problem is that whether $g:R^n \rightarrow R$ defined as $g(y):=\|[y]^+\|$ is concave. And if $g$ is concave then automatically $f$ will be concave.
Could anyone please share some thoughts?

Answer 1

This is a good question, and in general I don't know if an answer is known. I can recall the following cases in which $f$ is actually convex:

• if $X$ is a vector subspace, then the projection operator $[\cdot]^+$ is linear, so $f$ is convex as an affine transformation of $\|\cdot\|_2$.

• if $X$ is a cone, then a standard result (e.g., in Bauschke/Combettes' book on Convex Analysis and monotone operator theory in Hilbert spaces) states that $g$ is convex, so $f$ would retain convexity if $x=0$ as well.